Conduction to Displacement Current Ratio

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SUMMARY

The discussion focuses on calculating the ratio of conduction current density to displacement current density in copper when subjected to an alternating electric field. Using the conductivity value of copper, σ = 5.8 x 107 S/m, and a frequency of 1 megacycle/sec (1 MHz), the relationship is established as Jc/Jd = σ / (ωε). The derived formula indicates that the ratio is directly proportional to the conductivity and inversely proportional to the product of angular frequency and permittivity.

PREREQUISITES
  • Understanding of electric current density (J)
  • Knowledge of alternating electric fields and their representation (E0eiωt)
  • Familiarity with the concepts of conductivity (σ) and permittivity (ε)
  • Basic grasp of angular frequency (ω) in the context of AC circuits
NEXT STEPS
  • Study the derivation of current density equations in AC circuits
  • Learn about the physical significance of permittivity in materials
  • Explore the implications of frequency on conduction and displacement currents
  • Investigate the behavior of different materials under alternating electric fields
USEFUL FOR

Electrical engineers, physics students, and professionals working with electromagnetic theory and AC circuit analysis will benefit from this discussion.

atomicpedals
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Homework Statement



For copper we find that J=\sigmaE , where \sigma has the value 5.8x10^7 mksc units. If an alternating electric field represented by E0ei\omegat exists within the copper. Find the ratio of the conduction to the displacement current density at a frequency of 1megacycle/sec.

Homework Equations



J=\sigma dE0/dt

The Attempt at a Solution



So I'm hoping this is a straight forward as I'm making it out to be, minus my own issues with the units. I think it follows like this, but it seems too easy!

Please excuse my tex failures below...

J=\sigma dE0/dt = i\omega\sigmaE0
\omega\sigmaE0=E0/J = 1.04x10^12
 
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Got it, I think, Jc/Jd = \sigma / \omega \epsilon
 

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